A) \[{{C}_{2}}{{H}_{7}}{{N}_{2}}\]
B) \[C{{H}_{5}}N\]
C) \[C{{H}_{4}}N\]
D) \[{{C}_{2}}{{H}_{7}}N\]
Correct Answer: C
Solution :
As the sum of the percentage of C, H & N is 100. Thus it does not contains O atom.| Table for empirical formula | ||||
| Element | % | AT./wt. | Rel. Number | Ratio |
| C | 40.00 | 12 | \[\frac{40}{12}=3.66\] | \[\frac{3.66}{3.33}=1.09\tilde{\ }1\] |
| H | 13.33 | 1 | \[\frac{13.33}{1}=13.33\] | \[\frac{13.33}{3.33}=4\] |
| N | 46.67 | 14 | \[\frac{46.67}{14}=3.33\] | \[\frac{3.33}{3.33}=1\] |
| Hence empirical formula=\[C{{H}_{4}}N\] | ||||
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